Answer:
Choice C. [tex]84\, x + C[/tex].
Step-by-step explanation:
Consider the power rule for integration. Let [tex]n[/tex] be a real number that is not equal to [tex](-1)[/tex]. The power rule for integration states that:
[tex]\displaystyle \int x^{n}\, d x = \frac{1}{n + 1}\, x^{n+ 1} + C[/tex],
How could this rule apply to this question, since there's apparently no [tex]x[/tex] (or its powers) in the integrand? Keep in mind that [tex]x^{0} = 1[/tex] for all real (and particularly non-zero) values of [tex]x[/tex]. In other words, the integrand [tex]84[/tex] is equal to [tex]84\, x^0[/tex]. The integral becomes:
[tex]\displaystyle \int 84\, x^{0}\, dx[/tex].
The constant can be moved outside the integral sign. Therefore:
[tex]\displaystyle \int 84\, x^{0}\, dx= 84 \int x^{0}\, dx[/tex].
Now that resembles the power rule. In particular, [tex]n = 0[/tex], such that [tex]n + 1 = 1[/tex]. By the power rule:
[tex]\begin{aligned}84 \int x^{0}\, dx = 84\, \left(\frac{1}{1}\, x^{1} + C\right) = 84\, x + 84\, C\end{aligned}[/tex].
The non-zero constant in front of [tex]C[/tex] can be ignored (where [tex]C[/tex] represents the constant of integration.) Therefore:
[tex]\displaystyle \int 84\, dx = 84\, x + C[/tex].
If a is a constant then it's inetgration is
[tex]\boxed{\sf \displaystyel\int adx=ax+C}[/tex]
[tex]\\ \rm\Rrightarrow \displaystyle\int 84dx[/tex]
[tex]\\ \rm\Rrightarrow 84x+C[/tex]
Option C
